Design of Control Systems

 Question 1
The transfer function of a phase lead compensator is given by
$D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}$
The frequency (in rad/sec), at which $\angle D(j\omega )$ is maximum, is
 A $\sqrt{\frac{3}{T^2}}$ B $\sqrt{\frac{1}{3T^2}}$ C $\sqrt{3T}$ D $\sqrt{3T^3}$
GATE EE 2019   Control Systems
Question 1 Explanation:
$T(s)=\frac{1+3Ts}{1+Ts}$
Frequency at which $\angle T(j\omega )$ is maximum,
$(\omega _m)=\frac{1}{T\sqrt{\alpha }}$
$\alpha =\frac{1}{1/3}=3$
$\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}$
 Question 2
The transfer function C(s) of a compensator is given below.
$C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}$
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
 A $0.1 \lt \omega \lt 1$ B $1 \lt \omega \lt 10$ C $10 \lt \omega \lt 100$ D $\omega \gt 100$
GATE EE 2017-SET-2   Control Systems
Question 2 Explanation:
Pole zero diagram of compensator transfer function is shown below.

Maximum phase lead is between 0.1 and 1.
$0.1 \lt \omega \lt 1$
 Question 3
For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.
 A 0.15 B 0.32 C 0.66 D 0.92
GATE EE 2016-SET-2   Control Systems
Question 3 Explanation:
Assuming
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
 Question 4
The transfer function of a compensator is given as
$G_c(s)=\frac{s+a}{s+b}$
The phase of the above lead compensator is maximum at
 A $\sqrt{2}$ rad/s B $\sqrt{3}$ rad/s C $\sqrt{6}$ rad/s D $1/ \sqrt{3}$ rad/s
GATE EE 2012   Control Systems
Question 4 Explanation:
$\omega _m=\sqrt{\omega _{c1} \times \omega _{c2}}$
$\;\;=\sqrt{1 times 2}=\sqrt{2}$ rad/sec.
 Question 5
The transfer function of a compensator is given as
$G_c(s)=\frac{s+a}{s+b}$
$G_c(s)$ is a lead compensator if
 A a = 1, b = 2 B a = 3, b = 2 C a =- 3, b =- 1 D a = 3, b = 1
GATE EE 2012   Control Systems
Question 5 Explanation:
$G_c(s)=\frac{s+a}{s+b}$
$G_c(j\omega )=\frac{j\omega +a}{j\omega +b}$
$\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}$
$\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )$
For $G_c(s)$ to be a lead compensator
$\angle G_c(j\omega ) \gt 0$
$\frac{\omega }{a} \gt \frac{\omega }{b}$
$\Rightarrow b \gt a$
Option (A) Satisfies the above equation.
 Question 6
The transfer functions of two compensators are given below :

$C_{1}=\frac{10(s+1)}{(s+10)},C_{2}=\frac{s+10}{10(s+1)}$

Which one of the following statements is correct ?
 A $C_{1}$ is a lead compensator and $C_{2}$ is a lag compensator B $C_{1}$ is a lag compensator and $C_{2}$ is a lead compensator C Both $C_{1}\; and \; C_2$ are lead compensator D Both $C_{1}\; and \; C_2$ are lag compensator
GATE EE 2008   Control Systems
Question 6 Explanation:
$C_1=\frac{10(s+1)}{(s+10)}$
Zero at s=-1
Pole at s=-10

As zero is closer origin, zero dominates pole. Hence $C_1$ is lead compensator.
$C_2=\frac{s+10}{10(s+1)}$
Zero at s=-10
Pole at s=-1

As pole is closer to origin, pole dominates zero. Hence $C_2$ is lag compensator.
 Question 7
The system $\frac{900}{s(s+1)(s+9)}$ is to be compensated such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45$^{\circ}$ phase margin. To achieve this, one may use
 A a lag compensator that provides an attenuation of 20 dB and a phase lag of 45$^{\circ}$ at the frequency of 3 $\sqrt{3}$ rad/s B a lead compensator that provides an amplification of 20 dB and a phase lead of 45$^{\circ}$ at the frequency of 3 rad/s C a lag-lead compensator that provides an attenuation of 20 dB and phase lag of 45$^{\circ}$ at the frequency of $\sqrt{3}$ rad/s D a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45$^{\circ}$ at the frequency of 3 rad/s
GATE EE 2007   Control Systems
Question 7 Explanation:
Let uncompensated syatem,
$T(s)=\frac{900}{s(s+1)(s+9)}$
Phase crossover frequency of uncompensated system $=(\omega _{pc})$, at this frequency phase of $T(j \omega )$ is $-180^{\circ}$
Put $s=j\omega \text{ in } T(s)$
\begin{aligned} T(j\omega )&=\frac{900}{j\omega (j\omega +1)(j\omega +9)} \\ \angle T(j\omega )&=-90^{\circ}-tan ^{-1}\omega -tan^{-1}\left ( \frac{\omega }{9} \right ) \\ \text{At, } & (\omega _{pc})_1,\angle T(j\omega )=-180^{\circ}\\ -180^{\circ}&=-90^{\circ}-tan^{-1}(\omega _{pc})_1 -tan^{-1}\left ( \frac{(\omega _{pc})_1}{9} \right )\\ -90^{\circ}&=tan^{-1}\frac{(\omega _{pc})_1+\frac{(\omega _{pc})_1}{9}}{1-\frac{(\omega _{pc})^2_1}{9}} \\ \Rightarrow 1-\frac{(\omega _{pc})^2_1}{9}&=0 \\ \Rightarrow (\omega _{pc})_1&= 3 \text{ rad/sec} \end{aligned}
Gain cross frequency of compensated system, $(\omega _{gc})_2$= phase cross frequency of uncompensated system, $(\omega _{pc})_1$
$\Rightarrow \;\;(\omega _{gc})_2=(\omega _{gc})_1=\; 3$ rad/sec
Phase-margin $=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{gc})_2}$
$\Rightarrow \; \; \; 45^{\circ}=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{pc})_2}$
At $(\omega _{gc})_2=3$ rad/sec,
phase angle of $\angle T(j\omega )$ is $-135^{\circ}$ and phase of uncompensated systen is $-180^{\circ}$ at 3 rad/sec.
Therefore, the compensator provides phase lead of $45^{\circ}$ at the frequency of 3 rad/sec.
Let X dB is the gain provided by the compensator, so at gain cross frequency, $|T(j\omega )|_{com}$=1 or 0 dB. Gain of uncompensated system $=|T(j\omega )|_{un-com} =\frac{100}{\omega \sqrt{1+\omega ^2}\sqrt{1+\left ( \frac{\omega }{9} \right )^2}}$
$|T(j\omega )|_{un-com}$ in dB
$\;=40-20 \log \omega -20 \log \sqrt{1+\omega ^2} -20 \log \sqrt{1+\left ( \frac{\omega }{9} \right )^2}$
Gain of compensated system,
$|T(j\omega )|_{com}=X+|T(j\omega )_{un-com}|$
$|T(j\omega )|_{com}$ must be zero at gain cross frequency $(\omega _{gc})_2$
$|T(j\omega_{gc} )_2|_{com}=X+40-20 \log (\omega _{gc})_2 -20 \log \sqrt{1+(\omega _{gc})^2_2}-20 \log \sqrt{1+\frac{(\omega _{gc})^2}{9^2}}=0$
$X+40-20 \log 3-20 \log \sqrt{1+3^2}-20 \log \sqrt{1+\left ( \frac{3}{9} \right )^2}=0$
$X=-20dB$
So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.
 Question 8
A lead compensator used for a closed loop controller has the following transfer function $\frac{K(1+\frac{s}{a})}{1+\frac{s}{b}}$
 A $a \lt b$ B $a \gt b$ C $a \gt Kb$ D $a \lt Kb$
Transfer function$=\frac{k\left ( 1+\frac{s}{a} \right )}{\left ( 1+\frac{s}{b} \right )}$
So, from pole-zero configuration of the compensator $a \lt b$.